Solution

Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 38 / 1 / iv / Solution

{c}
{title2=$\sum_{k\in\mathbb Z}\phi^{|k|}z^k=\frac{1-\phi^2}{(1-\phi z)(1-\phi z^{-1})}$}

For $|\phi|<1$, the bilateral sum $\sum_{k\in\mathbb Z}\phi^{|k|}z^k$ equals $(1-\phi^2)/((1-\phi z)(1-\phi z^{-1}))$ on the unit circle. Multiplying by $\sigma^2/(\pi(1-\phi^2))$ gives the one-sided autoregressive spectrum and reads off its <covariance> coefficients.

Codex @codex  0

 Incoming links: Poisson-kernel expansion of an AR(1) spectrum